3×8854×1012×1.381×1023×273000084 1013×10 3.00094 =3.097×1040J1·C2·m a= a 2.783×10-30m E E+23E。kT 3.097×10×1013×103×10 =0.002293 3×8854×1012×1.381×10-23×373 1+2×0.00293 =1006895=1.0069 1-0.002293 4118 人o RT 3kT T=273K 3×8854×1042×8.314×273000569 6023×1023×101325×105( 3.00569 3×1.381×1023×273 T=373K, 3×8854×10-2×8.314×373000569 6023×1023×101325×10 3.00569 41l8=a+ 3×1.381×1023×373 32591×1039=a+8.8478×1042 25525×1039=a+64758×10192 解之得=5448×103C·m= 5 -12 -23 1.013 10 3 8.854 10 1.381 10 273       ( 3.00094 0.00084 )J-1·C 2·m2 = 3.097×10-40J -1·C 2·m2 ’= 4π 0   = 2.783×10-30m3 (2) 2 1 + −   = kT p 3 0   = 3 8.854 10 1.381 10 373 3.097 10 1.013 10 10 -12 -23 -40 5          = 0.002293  = 1- 0.002293 1+ 2 0.00293 = 1.006895 = 1.0069 4118  M m ( 2 1 + −   ) = 0 3 NA (  + 3kT 2  )  M m = p RT N p RT A 3 0  ( 2 1 + −   ) =  + 3kT 2  T = 273Ｋ, 23 5 -12 6.023 10 1.01325 10 3 8.854 10 8.314 273        ( 3.00569 0.00569 ) =  + 3 1.381 10 273 23 2     T = 373 Ｋ, 23 5 -12 6.023 10 1.01325 10 3 8.854 10 8.314 373        ( 3.00569 0.00569 ) 4118=  + 3 1.381 10 373 23 2     3.2591×10-39 =  + 8.8478×1019 2 2.5525×10-39 =  + 6.4758×1019 2 解之得  = 5.448×10-30C·m